Let $\Delta ABC$ be a triangle with angle $\angle CAB=60^{\circ}$, let $D$ be the intersection point of the angle bisector at $A$ and the side $BC$, and let $r_B,r_C,r$ be the respective radii of the incircles of $ABD$, $ADC$, $ABC$. Let $b$ and $c$ be the lengths of sides $AC$ and $AB$ of the triangle. Prove that \[ \frac{1}{r_B} +\frac{1}{r_C} ~=~ 2\cdot\left( \frac1r +\frac1b +\frac1c\right)\]
Problem
Source: Mediterranean Mathematical Olympiad 2019 P1 MMC
Tags: geometry, incircle, inradius
paperinikk
24.07.2019 00:05
$r=\frac{A_{ABC}}{p_{ABC}}=\frac{bc\sin 60}{a+b+c}$. It is well-known that $BD=\frac{ac}{b+c}$ and $DC=\frac{ab}{b+c}$ From Stewart we get $AD=\frac{\sqrt{bc((b+c)^2-a^2)}}{b+c}$ $r_B=\frac{A_{ABD}}{p_{ABD}}=\frac{\frac{1}{2}\frac{\sqrt{bc((b+c)^2-a^2)}}{b+c}c\sin 30}{\frac{1}{2}\left(c+\frac{ac}{b+c}+\frac{\sqrt{bc((b+c)^2-a^2)}}{b+c}\right)}=\frac{\sqrt{bc((b+c)^2-a^2)}c\sin 30}{c(b+c)+ac+\sqrt{bc((b+c)^2-a^2)}}$ $r_C=\frac{\sqrt{bc((b+c)^2-a^2)}b\sin 30}{b(b+c)+ab+\sqrt{bc((b+c)^2-a^2)}}$ similarly The equation becomes $$2\frac{c(b+c)+ac+\sqrt{bc((b+c)^2-a^2)}}{c\sqrt{bc((b+c)^2-a^2)}}+2\frac{b(b+c)+ab+\sqrt{bc((b+c)^2-a^2)}}{b\sqrt{bc((b+c)^2-a^2)}}=2\frac{a+b+c+(b+c)\sin 60}{bc\sin 60}$$ $$\frac{2bc(a+b+c)+(b+c)\sqrt{bc((b+c)^2-a^2)}}{bc\sqrt{bc((b+c)^2-a^2)}}=\frac{a+b+c+(b+c)\sin 60}{bc\sin 60}$$ $$ 2\sin 60 bc(a+b+c)+(b+c)\sin 60\sqrt{bc((b+c)^2-a^2)}=(a+b+c)\sqrt{bc((b+c)^2-a^2)}+(b+c)\sin 60\sqrt{bc((b+c)^2-a^2)}$$ After simplifying it remains to prove that $$\sqrt{3}bc=\sqrt{bc((b+c)^2-a^2)}$$ $a^2=b^2+c^2-2bc\cos 60=b^2+c^2-bc$ hence $\sqrt{bc((b+c)^2-a^2)}=\sqrt{bc(b^2+c^2+2bc-b^2-c^2+bc)}=\sqrt{3}bc$ This completes the proof
Panashe
28.03.2020 12:41
How can I save this page as a pdf file?
BestChoice123
28.03.2020 13:09
Panashe wrote: How can I save this page as a pdf file? Here
